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Let $x > 0$ be a real number. Prove that $x + \dfrac1{4x} \ge 1$.

I don't know where to begin with this question, I was hoping someone could help me out with this.

George Gronge
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    Instead of proving what you're asked, try to prove something equivalent, namely $4x^2-4x+1\ge 0$. – Git Gud Oct 09 '13 at 18:19
  • multiply both sides by x and try solving the equality. You shouldn't find any solution and since the to-be quadratic polynomial is in a U-shape you're done. – user88595 Oct 09 '13 at 18:20
  • @user88595 : If you do it that way, you need to treat the case $x>0$ and the case $x<0$ separately. In the second case, "$\ge$" will change to "$\le$". ${}\qquad{}$ – Michael Hardy Oct 09 '13 at 18:44
  • @MichaelHardy : The question states that $x>0$ so no need to worry about it. – user88595 Oct 09 '13 at 19:03
  • George: Please don't delete the substance of your question after receiving answers: It makes those answers seem irrelevant. Also, we encourage users to "accept" an answer that they find helpful. You can only accept one answer for each question you ask. To accept an answer, just click on the $\large \checkmark$ to the left of the answer you'd like to accept. It turns green when you click on it, and you receive $2$ reputation points for each answer you accept. – amWhy Oct 12 '13 at 14:28

2 Answers2

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HINT:

As $x>0$ and so is $\displaystyle \frac1{4x}$ use A.M. $\ge $ G.M.

lab bhattacharjee
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I'm going to approach this a bit differently from the most usual algorithms. Suppose we view $$ x+\frac{1}{4x}\tag1 $$ as $a^2+b^2$, and we seek a term that will be $-2ab$ to add between them, so that $a^2-2ab+b^2$ can be factored as $(a-b)^2$. We have $a=\sqrt{x}$ and $b = \dfrac{1}{2\sqrt{x}}$, so $2ab=1$. Then $$ x+\frac{1}{4x} = \left(x - 1 +\frac{1}{4x} \right) + 1 = \left(\sqrt{x}-\frac{1}{2\sqrt{x}}\right)^2+1 $$ Clearly this expression is $\ge 1$ and is $=1$ only if $x=1/2$. But all this works only if $x>0$. If $x<0$ then we can't take the square root, or if we allow imaginary numbers we can't conclude that the square is nonnegative. If $x=0$ then $1/(4x)$ is undefined.

So the set of values of $x$ for which the expression in $(1)$ above is $\ge 1$ is just the set of all positive numbers.

Michael Hardy
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